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enot [183]
3 years ago
10

Zoey ran a total of 1000m. 20 seconds into her run she had traveled 100 meters. What was her speed at 100 meters?

Physics
1 answer:
leva [86]3 years ago
7 0
Let's assume that Zoey ran at a constant speed. we can use the equation,
 d = st
where, d = distance, s = speed, and t = time taken.

By rearranging,
 s = d/t
Zoey had travelled 100 m in 20 seconds.
Hence, s = 100 m / 20 s = 5 m/s

therefore Zoey's speed at 100 m is 5 m/s
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(a)
Marta_Voda [28]

a) The momentum of the coconut is 3 kg m/s

b) At first, the air resistance is negligible, so the coconut accelerates due to the force of gravity

c) The coconut reaches its terminal velocity

Explanation:

a)

The momentum of an object is given by the equation

p=mv

where

m is the mass of the object

v is its velocity

For the coconut in this problem, we have:

m = 1.5 kg (mass)

v = 2 m/s (velocity)

Therefore, its momentum is

p=(1.5)(2)=3 kg m/s

B)

There are only two forces acting on the coconut during its fall:

  • The force of gravity, of magnitude mg (m= mass of the coconut, g = acceleration of gravity), acting downward
  • The air resistance, acting upward, whose magnitude is proportional to the speed of the coconut

During the first momentums of the fall, the speed of the coconut is still low, so the air resistance is mostly negligible, and therefore only the force of gravity is acting on the coconut. Since this force is constant, it means that the acceleration of the coconut is constant: therefore, its velocity keeps increasing during the fall, and the coconut speeds up.

C)

If the tree is very tall, the fall of the coconut lasts long, and the  speed of the coconut keeps increasing. Since the air resistance is proportional to the speed, this means that at some point, the air resistance is no longer negligible, and it starts to have some effect on the fall of the coconut. In particular, at a certain point, the air resistance will become equal (in magnitude) to the force of gravity (but opposite in direction): this means that  from this point, the acceleration of the coconut will be zero, and therefore the coconut will continue its motion at constant velocity. This velocity is called terminal velocity, and it occurs when the force of gravity is equal to the air resistance:

mg = F_r

where F_r is the air resistance.

Learn more about forces and weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

8 0
3 years ago
What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s
Ksivusya [100]

The net force is 270 N

Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 90.0 kg

a=3.0 m/s^2

Substituting, we find the net force on the object:

F=(90.0)(3.0)=270 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

3 0
3 years ago
A stone is dropped into a well. The sound of the splash is heard 3.5 seconds later. What is the depth of the well? Take the spee
Naddika [18.5K]

Answer:

The depth of the well, s = 54.66 m

Given:

time, t = 3.5 s

speed of sound in air, v = 343 m/s

Solution:

By using second equation of motion for the distance traveled by the stone when dropped into a well:

s = ut +\frac{1}{2}at^{2}

Since, the stone is dropped, its initial velocity, u = 0 m/s

and acceleration is due to gravity only, the above eqn can be written as:

s = \frac{1}{2}gt'^{2}

s = \frac{1}{2}9.8t^{2} = 4.9t'^{2}                     (1)

Now, when the sound inside the well travels back, the distance covered,s is given by:

s = v\times t''

s = 343\times t''                                              (2)

Now, total time taken by the sound to travel:

t = t' + t''

t'' = 3.5 - t'                                                                        (3)

Using eqn (2) and (3):

s = 343(3.5 - t')                                                                 (4)

from eqn (1) and (4):

4.9t'^{2} = 343(3.5 - t')

4.9t'^{2} + 343t' - 1200.5 = 0

Solving the above quadratic eqn:

t' = 3.34 s

Now, substituting t' = 3.34 s in eqn (2)

s = 54.66 m

3 0
3 years ago
A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc
Over [174]

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v is maximum tangential speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s

So, the maximum tangential speed is 11.7 m/s.

8 0
3 years ago
If you toast a piece of bread then spread butter onto the warm piece of toast, what would be the physical change occurring? What
slamgirl [31]
Wouldn’t it be from a solid to a liquid
8 0
2 years ago
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